Question

The equivalent weight of lodine in the
reaction
2Na2S2O3 +12 + 2Nal+Na2S406 is
(M = mol, wt]​

Answer 1

Answer:

2Na₂S₂O₃ + I₂ → Na₂S₄O₆ 2NaI

oxidation of S in S₂O₃⁻² = 2

oxidation number of S in S₄O₆⁻² = 5/2

again, 2S₂O₃⁻² → S₄O₆⁻²

for 2 moles of S₂O₃⁻²

change in oxidation number= 4x5/2 - 2x2x2

= 2

for 1 mole it'll be = 2/2 = 1

∴ equivalent mass of Na₂S₂O₃

= molar mass / change in oxi. no. = 158/1 = 158

l hope u like this answer please marks as brainlest